1.4.3 空间向量的应用--距离问题
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模块导图
![](https://cdn.nlark.com/yuque/0/2022/png/21447551/1642752442439-1eb83bed-ef7e-4033-b351-64b946b3d977.png )
知识剖析
利用空间向量法求距离问题
(1)点\(A、B\)间的距离
(2)点\(Q\)到直线\(l\) 距离
若\(Q\)为直线\(l\)外的一点,\(P\)在直线上,\(\vec{a}\)为直线\(l\)的方向向量,\(\vec{b}=\overrightarrow{P Q}\),
则点\(Q\)到直线\(l\)距离为\(d=\dfrac{1}{|\vec{a}|} \sqrt{(|\vec{a}||\vec{b}|)^{2}-(\vec{a} \cdot \vec{b})^{2}}\).
公式推导
如图,\(d=|\vec{b}| \sin \theta=|\vec{b}| \sqrt{1-\cos ^{2} \theta}\)\(=|\vec{b}| \sqrt{1-\left(\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)^{2}}=\dfrac{1}{|\vec{a}|} \sqrt{(|\vec{a}||\vec{b}|)^{2}-(\vec{a} \cdot \vec{b})^{2}}\)
(3)点\(Q\)到平面\(α\)的距离
若点\(Q\)为平面\(α\)外一点,点\(M\)为平面\(α\)内任一点,平面\(α\)的法向量为\(\vec{n}\),则\(Q\)到平面\(α\)的距离就等于\(\overrightarrow{M Q}\)在法向量\(\vec{n}\)方向上的投影的绝对值,即\(d=\dfrac{|\vec{n} \cdot \overrightarrow{M Q}|}{|\vec{n}|}\).
公式推导
如图,\(d=|\overrightarrow{M Q}| \sin \alpha=|\overrightarrow{M Q}||\cos \langle\vec{n}, \overrightarrow{M Q}\rangle|\)\(=|\overrightarrow{M Q}| \cdot \dfrac{|\vec{n} \cdot \overrightarrow{M Q}|}{|\vec{n}||\overrightarrow{M Q}|}=\dfrac{|\vec{n} \cdot \overrightarrow{M Q}|}{|\vec{n}|}\).
(4)直线\(a\)平面\(α\)之间的距离
当一条直线和一个平面平行时,直线上的各点到平面的距离相等.由此可知,直线到平面的距离可转化为求直线上任一点到平面的距离,即转化为点面距离.
(5) 平面间的距离
利用两平行平面间的距离处处相等,可将两平行平面间的距离转化为求点面距离.
经典例题
【题型一】点到点的距离
【典题1】正方体\(ABCD-A_1 B_1 C_1 D_1\)的棱长为\(1\),动点\(M\)在线段\(CC_1\)上,动点\(P\)在平面\(A_1 B_1 C_1 D_1\)上,且\(AP⊥\)平面\(MBD_1\).
(1)当点\(M\)与点\(C\)重合时,线段\(AP\)的长度为\(\underline{\quad \quad}\);
(2)线段\(AP\)长度的最小值为\(\underline{\quad \quad}\).
【解析】(1)以\(D\)为原点,\(DA\)为\(x\)轴,\(DC\)为\(y\)轴,\(DD_1\)为\(z\)轴,建立空间直角坐标系,
![](https://img2023.cnblogs.com/blog/1943255/202305/1943255-20230519155323221-172731538.png)
设\(P(a ,b ,1)\),当点\(M\)与\(C\)重合时,\(M(0 ,1 ,0)\),\(B(1 ,1 ,0)\),\(D_1 (0 ,0 ,1)\),\(A(1 ,0 ,0)\)
\(\overrightarrow{A P}=(a-1, b, 1)\),\(\overrightarrow{B D_{1}}=(-1,-1,1)\),\(\overrightarrow{M D_{1}}=(0,-1,1)\),
\(∵AP⊥\)平面\(MBD_1\).
\(\therefore\left\{\begin{array}{l}
\overrightarrow{A P} \cdot \overrightarrow{B D_{1}}=1-a-b+1=0 \\
\overrightarrow{A P} \cdot \overrightarrow{M D_{1}}=-b+1=0
\end{array}\right.\),解得\(a=1\),\(b=1\),
\(\therefore \overrightarrow{A P}=(0,1,1)\),
\(∴\)线段\(AP\)的长度为\(|\overrightarrow{A P}|=\sqrt{0+1+1}=\sqrt{2}\).
(2)设\(CM=t∈[0 ,1]\),则\(M(0 ,1 ,t)\),\(B(1 ,1 ,0)\),\(D_1 (0 ,0 ,1)\),
\(\overrightarrow{A P}=(a-1, b, 1)\),\(\overrightarrow{B D_{1}}=(-1,-1,1)\),\(\overrightarrow{M D_{1}}=(0,-1,1-t)\),
\(∵AP⊥\)平面\(MBD_1\).
\(\therefore\left\{\begin{array}{l}
\overrightarrow{A P} \cdot \overrightarrow{B D_{1}}=1-a-b+1=0 \\
\overrightarrow{A P} \cdot \overrightarrow{M D_{1}}=-b+1-t=0
\end{array}\right.\),
解得\(a=1+t\),\(b=1-t\),
\(∴\overrightarrow{A P}=(t, 1-t, 1)\),
\(\therefore|\overrightarrow{A P}|=\sqrt{t^{2}+(1-t)^{2}+1}=\sqrt{2\left(t-\dfrac{1}{2}\right)^{2}+\dfrac{3}{2}}\),
\(∴\)当\(t=\dfrac{1}{2}\),即\(M\)是\(CC_1\)中点时,线段\(AP\)长度取最小值为\(\dfrac{\sqrt{6}}{2}\).
【点拨】
① 线段\(AP\)的长度为\(|\overrightarrow{A P}|\),利用空间向量法使得几何问题“代数化”,较几何法更容易处理这动点问题;
② 本题的变化源头是“\(M\)的位置”,在第二问求\(AP\)长度的最小值,在引入参数中设\(CM=t\),较为合理.
巩固练习
1(★)已知\(M\)为\(z\)轴上一点,且点\(M\)到点\(A(-1 ,0 ,1)\)与点\((1 ,-3 ,2)\)的距离相等,则点\(M\)的坐标为\(\underline{\quad \quad}\).
2(★)已知空间直角坐标系\(O-xyz\)中有一点\(A(-1 ,-1 ,2)\),点\(B\)是\(xOy\)平面内的直线\(x+y=1\)上的动点,则\(A\),\(B\)两点的最短距离是\(\underline{\quad \quad}\).
3(★)如图,在空间直角坐标系中,有一棱长为\(2\)的正方体\(ABCD-A_1 B_1 C_1 D_1\),\(A_1 C\)的中点\(E\)到\(AB\)的中点\(F\)的距离为\(\underline{\quad \quad}\).
4(★)空间点\(A(x ,y ,z)\),\(O(0 ,0 ,0)\),\(B(\sqrt{2}, \sqrt{3}, 2)\),若\(|AO|=1\),则\(|AB|\)的最小值为\(\underline{\quad \quad}\).
答案
1.\((0 ,0 ,6)\)
2.\(\dfrac{\sqrt{34}}{2}\)
3.\(\sqrt{2}\)
4.\(2\)
【题型二】点到线的距离
【典题1】\(P\)为矩形\(ABCD\)所在平面外一点,\(PA⊥\)平面\(ABCD\),若已知\(AB=3\),\(AD=4\),\(PA=1\),则点\(P\)到\(BD\)的距离为\(\underline{\quad \quad}\).
【解析】\({\color{Red}{方法一 }}\)
\(∵\)矩形\(ABCD\)中,\(AB=3\),\(AD=4\),
\(\therefore B D=\sqrt{9+16}=5\),
过\(A\)作\(AE⊥BD\),交\(BD\)于\(E\),连结\(PE\),
\(∵PA⊥\)平面\(ABCD\),\(∴PA⊥BD\),
又\(AE⊥BD\)\(∴BD⊥\)平面\(PAE\),
\(∴PE⊥BD\),即\(PE\)是点\(P\)到\(BD\)的距离,
\(\because \dfrac{1}{2} \times A B \times A D=\dfrac{1}{2} \times B D \times A E\),
\(\therefore A E=\dfrac{A B \times A D}{B D}=\dfrac{12}{5}\),
\(\therefore P E=\sqrt{P A^{2}+E^{2}}=\sqrt{1+\dfrac{144}{25}}=\dfrac{13}{5}\),
\(∴\)点\(P\)到\(BD\)的距离为\(\dfrac{13}{5}\).
\({\color{Red}{方法二 }}\) 依题意可知,\(PA\)、\(AB\)、\(AD\)三线两两垂直,
如图建立空间直角坐标系
\(∴P(0 ,0 ,1)\),\(B(3 ,0 ,0)\),\(D(0 ,4 ,0)\),
\(\therefore \overrightarrow{B P}=(-3,0,1)\),\(\overrightarrow{B D}=(-3,4,0)\),
\(∴\)点\(P\)到\(BD\)的距离为\(d=\dfrac{1}{|\overrightarrow{B D}|} \sqrt{(|\overrightarrow{B D}||\overrightarrow{B P}|)^{2}-(\overrightarrow{B D} \cdot \overrightarrow{B P})^{2}}\)\(=\dfrac{1}{5} \cdot \sqrt{250-81}=\dfrac{13}{5}\)
【点拨】
① 方法一是几何法,找到点\(P\)到\(BD\)的距离\(PE\); 方法二是向量法,利用点到直线距离公式\(d=\dfrac{1}{|\overrightarrow{B D}|} \sqrt{(|\overrightarrow{B D}||\overrightarrow{B P}|)^{2}-(\overrightarrow{B D} \cdot \overrightarrow{B P})^{2}} \quad (*)\);
② 向量法中的公式\((*)\)有些复杂,不建议直接使用,还不如使用其推导方法
求点\(P\)到直线\(BD\)的距离
(1) 求出直线\(BD\)的方向向量\(\overrightarrow{B D}=(-3,4,0)\);
(2) 在直线\(BD\)上找一点\(B\),求出其与点\(P\)的向量\(\overrightarrow{B P}=(-3,0,1)\);
(3) 求两向量夹角余弦值,\(\cos <\overrightarrow{B P}, \overrightarrow{B D}>=\dfrac{\overrightarrow{B P} \cdot \overrightarrow{B D}}{|\overrightarrow{B P}||\overrightarrow{B D}|}=\dfrac{9}{5 \sqrt{10}}\);
(4) 求点\(P\)到\(BD\)的距离,\(d=|\overrightarrow{B P}| \sqrt{1-\cos ^{2}<\overrightarrow{B P}}, \overrightarrow{B D}>=\dfrac{13}{5}\).
巩固练习
1(★)已知直线\(l\)的方向向量为\(\vec{a}=(-1,0,1)\),点\(A(1 ,2 ,-1)\)在\(l\)上,则点\(P(2 ,-1 ,2)\)到\(l\)的距离为\(\underline{\quad \quad}\) .
2(★)已知直线\(l\)过定点\(A(2 ,3 ,1)\),且\(\vec{n}=(0,1,1)\)为其一个方向向量,则点\(P(4,3,2)\)到直线\(l\)的距离为\(\underline{\quad \quad}\) .
3(★★)已知\(A(0 ,0 ,2)\),\(B(1 ,0 ,2)\),\(C(0 ,2 ,0)\),则点\(A\)到直线\(BC\)的距离为\(\underline{\quad \quad}\).
答案
1.\(\sqrt{17}\)
2.\(\dfrac{3 \sqrt{2}}{2}\)
3.\(\dfrac{2 \sqrt{2}}{3}\)
【题型三】点到面的距离
【典题1】如图,四棱锥\(P-ABCD\)中,底面\(ABCD\)为菱形,\(∠ABC=60^°\),\(PA⊥\)平面\(ABCD\),\(AB=2\),\(P A=\dfrac{2 \sqrt{3}}{3}\),\(E\)为\(BC\)中点,\(F\)在棱\(PD\)上,\(AF⊥PD\),点\(B\)到平面\(AEF\)的距离为\(\underline{\quad \quad}\).
【解析】\(∵\)底面\(ABCD\)为菱形,\(∠ABC=60°\),\(E\)为\(BC\)中点,
\(∴AE⊥AD\),
又\(PA⊥\)平面\(ABCD\),\(∴PA⊥AD\),\(PA⊥AE\),
\(∴\)以\(A\)为原点,\(AE\)为\(x\)轴,\(AD\)为\(y\)轴,\(AP\)为\(z\)轴,建立空间直角坐标系,
\(∴A(0 ,0 ,0)\),\(B(\sqrt{3},-1,0)\)\(E(\sqrt{3}, 0,0)\),\(P\left(0,0, \dfrac{2 \sqrt{3}}{3}\right)\),\(D(0 ,2 ,0)\),
\(\therefore \overrightarrow{A B}=(\sqrt{3},-1,0)\),\(\overrightarrow{A E}=(\sqrt{3}, 0,0)\),
在\(Rt∆PAD\)中,易得\(∠D=30^°\),\(\therefore A F=\dfrac{A D}{2}=1\),
过点\(F\)作\(FH⊥AD\)交\(AD\)于\(H\),
易得\(∠1=30^°\),\(\therefore A H=\dfrac{1}{2}\),\(F H=\dfrac{\sqrt{3}}{2}\),
\(\therefore F\left(0, \dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)\),\(\overrightarrow{A F}=\left(0, \dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)\),
设平面\(AEF\)的法向量\(\vec{n}=(x, y, z)\),
则\(\left\{\begin{array}{l}
\vec{n} \cdot \overrightarrow{A E}=\sqrt{3} x=0 \\
\vec{n} \cdot \overrightarrow{A F}=\dfrac{1}{2} y+\dfrac{\sqrt{3}}{2} z=0
\end{array}\right.\),
取\(y=\sqrt{3}\),得\(\vec{n}=(0, \sqrt{3},-1)\),
\(∴\)点\(B\)到平面\(AEF\)的距离为\(d=\dfrac{|\vec{n} \cdot \overrightarrow{A B}|}{|\vec{n}|}=\dfrac{\sqrt{3}}{2}\).
【点拨】
① 求点\(F\)的坐标,解题中几何法较易求得,这需要审题中注意各量之间的关系;也可以用代数法求得,如下:
设\(F(0 ,b ,c)\),\(\overrightarrow{P F}=\lambda \overrightarrow{P D}\),
则\(\left(0, b, c-\dfrac{2 \sqrt{3}}{3}\right)=\left(0,2 \lambda,-\dfrac{2 \sqrt{3}}{3} \lambda\right)\),
解得\(b=2λ\),\(C=\dfrac{2 \sqrt{3}}{3}-\dfrac{2 \sqrt{3}}{3} \lambda\),
\(\therefore \overrightarrow{A F}=\left(0,2 \lambda, \dfrac{2 \sqrt{3}}{3}-\dfrac{2 \sqrt{3}}{3} \lambda\right)\),\(\overrightarrow{P D}=\left(0,2,-\dfrac{2 \sqrt{3}}{3}\right)\),
\(∵AF⊥PD\),\(\therefore \overrightarrow{A F} \cdot \overrightarrow{P D}=4 \lambda-\dfrac{4}{3}+\dfrac{4}{3} \lambda=0\),
解得\(\lambda=\dfrac{1}{4}\),\(\therefore F\left(0, \dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)\);
② 求点\(B\)到平面\(AEF\)的距离的解题步骤
(1) 求平面\(AEF\)的法向量\(\vec{n}=(0, \sqrt{3},-1)\);
(2) 在平面\(AEF\)内选一点\(A\),求其与点\(B\)的向量\(\overrightarrow{A B}=(\sqrt{3},-1,0)\);
(3) 利用公式\(d=\dfrac{|\vec{n} \cdot \overrightarrow{A B}|}{|\vec{n}|}\)(向量\(\vec{n}\)在法向量\(\vec{n}\)上的投影绝对值)求所求距离,\(d=\dfrac{|\vec{n} \cdot \overrightarrow{A B}|}{|\vec{n}|}=\dfrac{\sqrt{3}}{2}\).
【典题2】已知\(E\),\(F\)分别是正方形\(ABCD\)边\(AD\),\(AB\)的中点,\(EF\)交\(AC\)于\(P\),\(GC\)垂直于\(ABCD\)所在平面.
(1)求证:\(EF⊥\)平面\(GPC\).
(2)若\(AB=4\),\(GC=2\),求点\(B\)到平面\(EFG\)的距离.
【解析】(1)连接\(BD\)交\(AC\)于\(O\),
\(∵E\),\(F\)是正方形\(ABCD\)边\(AD\),\(AB\)的中点,\(AC⊥BD\),\(∴EF⊥AC\).
\(∵GC\)垂直于\(ABCD\)所在平面,\(EF⊂\)平面\(ABCD\)\(∴EF⊥GC\)
\(∵AC∩GC=C\),\(∴EF⊥\)平面\(GPC\).
(2) \({\color{Red}{方法一 \quad向量法 }}\)
建立空间直角坐标系\(C-xyz\),则\(G(0 ,0 ,2)\),\(E(4 ,2 ,0)\),\(F(2 ,4 ,0)\),\(B(4 ,0 ,0)\)
\(\therefore \overrightarrow{G E}=(4,2,-2)\),\(\overrightarrow{E F}=(-2,2,0)\)
设面\(GEF\)的法向量\(\vec{n}=(x, y, z)\),
则\(\overrightarrow{G E} \cdot \vec{n}=0\)且\(\overrightarrow{E F} \cdot \vec{n}=0\),
即\(4x+2y-2z=0\)且\(-2x+2y=0\)
取\(x=1\)时,可得\(\vec{n}=(1,1,3)\)
又\(∵\)向量\(\overrightarrow{B E}=(0,2,0)\)
则\(B\)到面\(GEF\)的距离\(d=\dfrac{|\vec{n} \cdot \overrightarrow{B E}|}{|\vec{n}|}=\dfrac{2 \sqrt{11}}{11}\).
\({\color{Red}{方法二 \quad等积法 }}\)
由题意可知\(P C=\dfrac{3}{4} A C=3 \sqrt{2}\),\(P G=\sqrt{P C^{2}+G C^{2}}=\sqrt{22}\),
\(\therefore S_{\Delta E F G}=\dfrac{1}{2} \times P G \times E F=2 \sqrt{11}\),
易得\(S_{\Delta E F B}=\dfrac{1}{2} \times A F \times E B=2\)
\(\therefore V_{B-E F G}=V_{G-E F B}\)
\(\therefore \dfrac{1}{3} \times h \times S_{\Delta E F G}=\dfrac{1}{3} \times G C \times S_{\Delta E F B}\)
\(\therefore h=\dfrac{G C \times S_{\Delta E F B}}{S_{\Delta E F G}}=\dfrac{2 \times 2}{2 \sqrt{11}}=\dfrac{2 \sqrt{11}}{11}\)
\({\color{Red}{方法三 \quad 间接法 }}\)
由题意可知\(P C=\dfrac{3}{4} A C=3 \sqrt{2}\),\(P G=\sqrt{P C^{2}+G C^{2}}=\sqrt{22}\),
\(∵PC=3OP\),
\(∴C\)到面\(GEF\)的距离是\(O\)到面\(GEF\)距离的\(3\)倍,
在\(∆GPC\)中,点\(C\)到边\(PG\)的高为\(CM\),
又\(∵EF⊥\)平面\(GPC\),\(∴CM⊥\)平面\(EFG\),
\(∴CM\)为\(C\)到面\(GEF\)距离,
在\(∆GPC\)中,可得\(P G \cdot C M=G C \cdot P C \Rightarrow C M=\dfrac{2 \times 3 \sqrt{2}}{\sqrt{22}}=\dfrac{6}{\sqrt{11}}\),
又\(BD∥EF\),可得\(BD∥\)平面\(GEF\),
可得\(B\)到面\(GEF\)的距离等于\(O\)到面\(GEF\)的距离\(\dfrac{1}{3} C M=\dfrac{2}{\sqrt{11}}=\dfrac{2 \sqrt{11}}{11}\).
【点拨】
求点\(A\)到平面\(α\)的距离方法有很多种,
① 直接法:若能确定点\(A\)到平面\(α\)的垂线段当然最好了!
② 向量法:若空间直角坐标系较容易建立,各关键点的坐标易求,可考虑向量法;本题中\(GC⊥\)平面\(ABCD\),矩形\(ABCD\)都是有利条件;
③等积法:当相关三棱锥的体积和侧面三角形的面积易求,可考虑等积法;本题中的\(V_{G-E F B}\)和\(S_{∆EFB}\)均易求;
④ 间接法:若存在过点\(A\)的直线\(l\)与平面\(α\)平行,可考虑能否在直线\(l\)上找到一点\(B\),而它到平面\(α\)的距离易求些;本题中“求\(B\)到面\(GEF\)的距离”转化为“求\(O\)到面\(GEF\)的距离”;
【典题3】如图在四棱锥\(P-ABCD\)中,侧面\(PAD⊥\)底面\(ABCD\),侧棱\(\text { 夌 } P A=P D=\sqrt{2}\),底面\(ABCD\)为直角梯形,其中\(BC∥AD\),\(AB⊥AD\),\(AD=2AB=2BC=2\),\(O\)为\(AD\)的中点.
(1)求证\(PO⊥\)平面\(ABCD\);
(2)求二面角\(C-PD-A\)夹角的正弦值;
(3)线段\(AD\)上是否存在\(Q\),使得它到平面\(PCD\)的距离为\(\dfrac{\sqrt{3}}{2}\)?若存在,求出\(\dfrac{A Q}{Q D}\)的值;若不存在,说明理由.
【解析】(1)证明\(∵PA=PD\),\(O\)为\(AD\)的中点,\(∴PO⊥AD\),
\(∵\)侧面\(PAD⊥\)底面\(ABCD\),侧面\(PAD∩\)底面\(ABCD=AD\),
\(∴PO⊥\)平面\(ABCD\).
(2)\(∵\)底面\(ABCD\)为直角梯形,其中\(BC∥AD\),\(AB⊥AD\),\(AD=2AB=2BC=2\),
\(∴OC⊥AD\),又\(PO⊥\)平面\(ABCD\),
\(∴\)以\(O\)为原点,\(OC\)为\(x\)轴,\(OD\)为\(y\)轴,\(OP\)为\(z\)轴,建立空间直角坐标系,
平面\(PAD\)的法向量\(\vec{m}=(1,0,0)\),
\(C(1 ,0 ,0)\),\(D(0 ,1 ,0)\),\(P(0 ,0 ,1)\),\(\overrightarrow{P C}=(1,0,-1)\),\(\overrightarrow{P D}=(0,1,-1)\),
设平面\(PCD\)的法向量\(\vec{n}=(x, y, z)\),
则\(\left\{\begin{array}{l}
\vec{n} \cdot \overrightarrow{P C}=x-z=0 \\
\vec{n} \cdot \overrightarrow{P D}=y-z=0
\end{array}\right.\),取\(x=1\),得\(\vec{n}=(1,1,1)\),
设二面角\(C-PD-A\)夹角为\(θ\),则\(\cos \theta=\dfrac{|\vec{m} \cdot \vec{n}|}{|\vec{m}| \cdot|\vec{n}|}=\dfrac{1}{\sqrt{3}}\),
\(\therefore \sin \theta=\sqrt{1-\left(\dfrac{1}{\sqrt{3}}\right)^{2}}=\dfrac{\sqrt{6}}{3}\),
\(∴\)二面角\(C-PD-A\)夹角的正弦值为\(\dfrac{\sqrt{6}}{3}\).
(3)设线段\(AD\)上存在\(Q(0 ,m ,0)\),\(m∈[-1 ,1]\),使得它到平面\(PCD\)的距离为\(\dfrac{\sqrt{3}}{2}\),
\(\therefore \overrightarrow{P Q}=(0, m,-1)\),
\(∴Q\)到平面\(PCD\)的距离\(d=\dfrac{|\overrightarrow{P Q} \cdot \vec{n}|}{|\vec{n}|}=\dfrac{|m-1|}{\sqrt{3}}=\dfrac{\sqrt{3}}{2}\),
解得\(m=-\dfrac{1}{2}\)或\(m=\dfrac{5}{2}\)(舍去),
\(\therefore Q\left(0,-\dfrac{1}{2}, 0\right)\),则\(\dfrac{A Q}{Q D}=\dfrac{\dfrac{1}{2}}{\dfrac{3}{2}}=\dfrac{1}{3}\).
【点拨】立体几何中问到是否“存在”,可利用“假设法”.
巩固练习
1(★★)在长方体\(ABCD-A_1 B_1 C_1 D_1\)中,\(AB=AD=2\),\(AA_1=1\),则点\(B\)到平面\(D_1 AC\)的距离等于\(\underline{\quad \quad}\).
2(★★)已知平面\(α\)的法向量为\(\vec{n}=(-2,-2,1)\),点\(A(x ,3 ,0)\)在平面\(α\)内,则点\(P(-2 ,1 ,4)\)到平面\(α\)的距离为\(\dfrac{10}{3}\),则\(x=\)\(\underline{\quad \quad}\).
3(★★)如图,\(ABCD\)是矩形,\(PD⊥\)平面\(ABCD\),\(PD=DC=a\),\(A D=\sqrt{2} a\),\(M\),\(N\)分别是\(AD\)、\(PB\)的中点,求点\(A\)到平面\(MNC\)的距离.
4(★★)如图,在长方体\(ABCD-A_1 B_1 C_1 D_1\)中,\(AD=AA_1=1\),\(AB=2\),点\(E\)在棱\(AB\)上移动.
(1)证明:\(D_1 E⊥A_1 D\);
(2)当\(E\)为\(AB\)的中点时,求点\(E\)到面\(ACD_1\)的距离.
5(★★★)已知三棱锥\(S-ABC\),满足\(SA\),\(SB\),\(SC\)两两垂直,且\(SA=SB=SC=2\),\(Q\)是三棱锥\(S-ABC\)外接球上一动点,求点\(Q\)到平面\(ABC\)的距离的最大值.
6(★★★)如图,在三棱锥\(S-ABC\)中,\(△ABC\)是边长为\(4\)的正三角形,\(S A=S C=2 \sqrt{2}\),\(O\),\(M\)分别为\(AC\)、\(AB\)的中点,\(SO⊥AB\).
(1)证明:\(SO⊥\)平面\(ABC\);
(2)求二面角\(S-CM-A\)的余弦值;
(3)求点\(B\)到平面\(SCM\)的距离.
参考答案
1.\(\dfrac{\sqrt{6}}{3}\)
2.\(-1\)或\(-11\)
3.\(\dfrac{a}{2}\)
4.\((1)\)略\(\text { (2) } \dfrac{1}{3}\)
5.\(\dfrac{4 \sqrt{3}}{3}\)
6.\((1)\)略\(\text { (2) } \dfrac{\sqrt{5}}{5}\)\(\text { (3) } \dfrac{4 \sqrt{5}}{5}\)
【题型四】线到面的距离
【典题1】如图,已知斜三棱柱\(ABC-A_1 B_1 C_1\),\(∠BCA=90°\),\(AC=BC=2\),\(A_1\)在底面\(ABC\)上的射影恰为\(AC\)的中点\(D\), 又知\(BA_1⊥AC_1\).
(1)求证:\(AC_1⊥\)平面\(A_1 BC\);
(2)求\(CC_1\)到平面\(A_1 AB\)的距离.
【解析】(1)\(∵A_1\)在底面\(ABC\)上的射影为\(AC\)的中点\(D\),
\(∴\)平面\(A_1 ACC_1⊥\)平面\(ABC\),
\(∵BC⊥AC\)且平面\(A_1 ACC_1∩\)平面\(ABC=AC\),
\(∴BC⊥\)平面\(A_1 ACC_1\),
\(∴BC⊥AC_1\),
\(∵AC_1⊥BA_1\)且\(BC∩BA_1=B\),
\(∴AC_1⊥\)平面\(A_1 BC\).
(2)如图所示,以\(C\)为坐标原点建立空间直角坐标系,
\(∵AC_1⊥\)平面\(A_1 BC\),\(∴AC_1⊥A_1 C\),
\(∴\)四边形\(A_1 ACC_1\)是菱形,
\(∵D\)是\(AC\)的中点,\(A_1 D⊥AC\)
\(∴∠A_1 AD=60°\),
\(∴A(2 ,0 ,0)\),\(A_{1}(1,0, \sqrt{3})\),\(B(0 ,2 ,0)\),\(C_{1}(-1,0, \sqrt{3})\),
\(\therefore \overrightarrow{A_{1} A}=(1,0,-\sqrt{3})\),\(\overrightarrow{A B}=(-2,2,0)\),
设平面\(A_1 AB\)的法向量\(\vec{n}=(x, y, z)\),
则\(\left\{\begin{array}{l}
\vec{n} \cdot \overrightarrow{A_{1} A}=0 \\
\vec{n} \cdot \overrightarrow{A B}=0
\end{array}\right.\),
\(\therefore\left\{\begin{array}{l}
x-\sqrt{3} z=0 \\
-2 x+2 y=0
\end{array}\right.\),\(\vec{n}=(\sqrt{3}, \sqrt{3}, 1)\),
\(\because \overrightarrow{C A_{1}}=(2,0,0)\),
\(∴C_1\)到平面\(A_1 AB\)的距离\(d=\dfrac{\left|\overrightarrow{C A_{1}} \cdot \vec{n}\right|}{|\vec{n}|}=\dfrac{2 \sqrt{21}}{7}\).
\(∵CC_1//AA_1\),\(AA_1⊂\)平面\(A_1 AB\),\(CC_1⊄\)平面\(A_1 AB\)
\(∴CC_1//\)平面\(A_1 AB\),
\(∴CC_1\)到平面\(A_1 AB\)的距离等于\(C_1\)到平面\(A_1 AB\)的距离\(\dfrac{2 \sqrt{21}}{7}\).
【点拨】直线到平面的距离问题可转化为点到平面的距离.
巩固练习
1(★★)如图,在棱长为\(1\)的正方体\(ABCD-A_1 B_1 C_1 D_1\)中,\(E\)为线段\(A_1 B_1\)的中点,\(F\)为线段\(AB\)的中点.求直线\(FC\)到平面\(AEC_1\)的距离;
2(★★)如图,在正方体\(ABCD-A_1 B_1 C_1 D_1中\),\(E\)为\(BB_1\)的中点.
(Ⅰ)证明:\(BC_1∥\)平面\(AD_1 E\);
(Ⅱ)求直线\(BC_1\)到平面\(AD_1 E\)的距离;
参考答案
1.\(\dfrac{\sqrt{6}}{6}\)
2.\((1)\)略\(\text { (2) } \dfrac{2}{3}\)
【题型五】面到面的距离
【典题1】正方体\(ABCD-A_1 B_1 C_1 D_1\)的棱长为\(a\),则平面\(AB_1 D_1\)与平面\(BDC_1\)的距离为\(\underline{\quad \quad}\).
【解析】\({\color{Red}{方法一 }}\) 连接\(A_1 C\),与面\(AB_1 D_1\)与面\(BC_1 D\)分别交于\(M\),\(N\).
\(∵CC_1⊥\)平面\(A_1 B_1 C_1 D_1\),\(∴CC_1⊥B_1 D_1\),
又\(∵A_1 C_1⊥B_1 D_1\),
\(∴B_1 D_1⊥\)平面\(A_1 C_1 C\)
\(∴B_1 D_1⊥A_1 C\),
同理可证\(AB_1⊥A_1 C\),
又\(B_1 D_1∩AB_1=B_1\),\(∴A_1 C⊥\)面\(AB_1 D_1\);
同理可证,\(A_1 C⊥\)面\(BC_1 D\).
\(∴MN\)为平面\(AB_1 D_1\)与平面\(BC_1 D\)的距离
\(∵△AB_1 D_1\)为正三角形,边长为\(\sqrt{2} a\),三棱锥\(A_1-AB_1 D_1\)为正三棱锥,
\(∴M\)为\(△AB_1 D_1\)的中心,\(M A=\dfrac{\sqrt{3}}{3} \times \sqrt{2} a=\dfrac{\sqrt{6}}{3} a\)
\(A_{1} M=\sqrt{A_{1} A^{2}-M A^{2}}=\dfrac{\sqrt{3}}{3} a\),
同理求出\(C N=A_{1} M=\dfrac{\sqrt{3}}{3} a\),
又\(A_{1} C=\sqrt{3} a\),
\(\therefore M N=A_{1} C-A_{1} M-C N=\dfrac{\sqrt{3}}{3} a\).
\({\color{Red}{方法二 }}\) 建立空间直角坐标系如图.
则\(A(a ,0 ,0)\),\(B(a ,a ,0)\),\(D(0 ,0 ,0)\),\(C_1 (0 ,a ,a)\),\(D_1 (0 ,0 ,a)\),\(B_1 (a ,a ,a)\),
\(\therefore \overrightarrow{A B_{1}}=(0, a, a)\),\(\overrightarrow{A D_{1}}=(-a, 0, a)\),\(\overrightarrow{B C_{1}}=(-a, 0, a)\),\(\overrightarrow{D C_{1}}=(0, a, a)\)
设\(\vec{n}=(x, y, z)\)为平面\(AB_1 D_1\)的法向量,
则\(\left\{\begin{array}{c}
\vec{n} \cdot \overrightarrow{A B_{1}}=a(y+z)=0 \\
\vec{n} \cdot \overrightarrow{A D_{1}}=a(-x+z)=0
\end{array}\right.\)得\(\left\{\begin{array}{c}
y=-z \\
x=z
\end{array}\right.\)
令\(z=1\),则\(\vec{n}=(1,-1,1)\)
\(\because \overrightarrow{A D_{1}} / / \overrightarrow{B C_{1}}\),\(\overrightarrow{A B_{1}} / / \overrightarrow{D C_{1}}\),
\(∴AD_1∥BC_1\),\(AB_1∥DC_1\),\(AD_1∩AB_1=A\),\(DC_1∩BC_1=C_1\),
\(∴\)平面\(AB_1 D_1∥\)平面\(BDC_1\)
\(∴\)平面\(AB_1 D_1\)与平面\(BDC_1\)的距离等于点\(C_1\)到平面\(AB_1 D_1\)的距离\(d\).
\(\because \overrightarrow{C_{1} B_{1}}=(a, 0,0)\),平面\(AB_1 D_1\)的法向量为\(\vec{n}=(1,-1,1)\),
\(\therefore d=\dfrac{\left|\overrightarrow{C_{1} B_{1}} \cdot \vec{n}\right|}{|\vec{n}|}=\dfrac{\sqrt{3}}{3} a\).
【点拨】
① 本题里,正方体中\(M\)、\(N\)把体对角线\(A_1 C\)三等分,即\(A_{1} M=M N=N C=\dfrac{A_{1} C}{3}=\dfrac{\sqrt{3}}{3} a\),这可作为一个结论记住;
② 面面间的距离问题可转化为点到面的距离.本题中平面\(AB_1 D_1∥\)平面\(BDC_1\), 它们间的距离转化为点\(C_1\)到平面\(AB_1 D_1\)的距离\(d\).
巩固练习
1(★★★)直四棱柱\(ABCD-A_1 B_1 C_1 D_1\)中,底面\(ABCD\)为正方形,边长为\(2\),侧棱\(A_1 A=3\),\(M\)、\(N\)分别为\(A_1 B_1\)、\(A_1 D_1\)的中点,\(E\)、\(F\)分别是\(B_1 C_1\)、\(C_1 D_1\)的中点.
(1)求证:平面\(AMN∥\)平面\(EFDB\);
(2)求平面\(AMN\)与平面\(EFDB\)的距离.
参考答案
1.\((1)\)略\(\text { (2) } \dfrac{6 \sqrt{19}}{19}\)